Hausdorff $2$-dimensional measure of $mathbf{R}$ [closed]
$begingroup$
I know that $mathcal{H}^2(mathbf{R}) = 0$, but what is any easy way to see it? I tried coming up with reasonable covers, but they all seem to give upper bounds that are too large.
measure-theory hausdorff-measure
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
$endgroup$
closed as off-topic by Xander Henderson, amWhy, Did, jgon, Shailesh Jan 9 at 2:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, amWhy, Did, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I know that $mathcal{H}^2(mathbf{R}) = 0$, but what is any easy way to see it? I tried coming up with reasonable covers, but they all seem to give upper bounds that are too large.
measure-theory hausdorff-measure
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
$endgroup$
closed as off-topic by Xander Henderson, amWhy, Did, jgon, Shailesh Jan 9 at 2:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, amWhy, Did, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I know that $mathcal{H}^2(mathbf{R}) = 0$, but what is any easy way to see it? I tried coming up with reasonable covers, but they all seem to give upper bounds that are too large.
measure-theory hausdorff-measure
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
$endgroup$
I know that $mathcal{H}^2(mathbf{R}) = 0$, but what is any easy way to see it? I tried coming up with reasonable covers, but they all seem to give upper bounds that are too large.
measure-theory hausdorff-measure
measure-theory hausdorff-measure
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
asked Jan 8 at 14:54
Drew BradyDrew Brady
672315
672315
closed as off-topic by Xander Henderson, amWhy, Did, jgon, Shailesh Jan 9 at 2:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, amWhy, Did, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Xander Henderson, amWhy, Did, jgon, Shailesh Jan 9 at 2:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Xander Henderson, amWhy, Did, jgon, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can cover by rectangles of small radius. Fix $delta>0$ and consider covering of $mathbb{R}$ by intervals of length $frac{delta}{n}$, where $nin mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $mathbb{R}$ and let ${I_n}$ be some enumeration of these intervals (the exact order does not matter). For $nin mathbb{N}$, consider a rectangle of height $delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $bigcuplimits_{n}^infty I_n$ covers $mathbb{R}$ with sets of diameter bounded above by $2delta$. However
$$
H^2_{2delta}(mathbb{R}) leq sumlimits_{n=1}^infty mathrm{diam}(I_n)^2 leq 2delta^2 sumlimits_{n=1}^infty frac{1}{n^2} leq 10 delta^2,
$$
hence $H^2(mathbb{R}) = limlimits_{delta to 0} H^2_delta(mathbb{R}) = 0$.
answered Jan 8 at 15:08
HaykHayk
2,2671214
$endgroup$
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can cover by rectangles of small radius. Fix $delta>0$ and consider covering of $mathbb{R}$ by intervals of length $frac{delta}{n}$, where $nin mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $mathbb{R}$ and let ${I_n}$ be some enumeration of these intervals (the exact order does not matter). For $nin mathbb{N}$, consider a rectangle of height $delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $bigcuplimits_{n}^infty I_n$ covers $mathbb{R}$ with sets of diameter bounded above by $2delta$. However
$$
H^2_{2delta}(mathbb{R}) leq sumlimits_{n=1}^infty mathrm{diam}(I_n)^2 leq 2delta^2 sumlimits_{n=1}^infty frac{1}{n^2} leq 10 delta^2,
$$
hence $H^2(mathbb{R}) = limlimits_{delta to 0} H^2_delta(mathbb{R}) = 0$.
answered Jan 8 at 15:08
HaykHayk
2,2671214
$endgroup$
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
|
show 1 more comment
$begingroup$
You can cover by rectangles of small radius. Fix $delta>0$ and consider covering of $mathbb{R}$ by intervals of length $frac{delta}{n}$, where $nin mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $mathbb{R}$ and let ${I_n}$ be some enumeration of these intervals (the exact order does not matter). For $nin mathbb{N}$, consider a rectangle of height $delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $bigcuplimits_{n}^infty I_n$ covers $mathbb{R}$ with sets of diameter bounded above by $2delta$. However
$$
H^2_{2delta}(mathbb{R}) leq sumlimits_{n=1}^infty mathrm{diam}(I_n)^2 leq 2delta^2 sumlimits_{n=1}^infty frac{1}{n^2} leq 10 delta^2,
$$
hence $H^2(mathbb{R}) = limlimits_{delta to 0} H^2_delta(mathbb{R}) = 0$.
answered Jan 8 at 15:08
HaykHayk
2,2671214
$endgroup$
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
|
show 1 more comment
$begingroup$
You can cover by rectangles of small radius. Fix $delta>0$ and consider covering of $mathbb{R}$ by intervals of length $frac{delta}{n}$, where $nin mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $mathbb{R}$ and let ${I_n}$ be some enumeration of these intervals (the exact order does not matter). For $nin mathbb{N}$, consider a rectangle of height $delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $bigcuplimits_{n}^infty I_n$ covers $mathbb{R}$ with sets of diameter bounded above by $2delta$. However
$$
H^2_{2delta}(mathbb{R}) leq sumlimits_{n=1}^infty mathrm{diam}(I_n)^2 leq 2delta^2 sumlimits_{n=1}^infty frac{1}{n^2} leq 10 delta^2,
$$
hence $H^2(mathbb{R}) = limlimits_{delta to 0} H^2_delta(mathbb{R}) = 0$.
answered Jan 8 at 15:08
HaykHayk
2,2671214
$endgroup$
You can cover by rectangles of small radius. Fix $delta>0$ and consider covering of $mathbb{R}$ by intervals of length $frac{delta}{n}$, where $nin mathbb{N}$. This is possible in view of the divergence of the harmonic series. Namely, ignoring $delta>0$ for a moment, start with $[0,1]$, then consider $[1, 1 + 1/2]$, $[1+1/2, 1+1/2+1/3]$, ... . Repeat the same construction for the negative axis. We get a countable cover of $mathbb{R}$ and let ${I_n}$ be some enumeration of these intervals (the exact order does not matter). For $nin mathbb{N}$, consider a rectangle of height $delta>0$ whose side length is equal to the length of $I_n$ and which contains the interval $I_n$. Then the union $bigcuplimits_{n}^infty I_n$ covers $mathbb{R}$ with sets of diameter bounded above by $2delta$. However
$$
H^2_{2delta}(mathbb{R}) leq sumlimits_{n=1}^infty mathrm{diam}(I_n)^2 leq 2delta^2 sumlimits_{n=1}^infty frac{1}{n^2} leq 10 delta^2,
$$
hence $H^2(mathbb{R}) = limlimits_{delta to 0} H^2_delta(mathbb{R}) = 0$.
answered Jan 8 at 15:08
HaykHayk
2,2671214
edited Jan 8 at 15:40
answered Jan 8 at 15:08
HaykHayk
2,2671214
answered Jan 8 at 15:08
HaykHayk
2,2671214
answered Jan 8 at 15:08
HaykHayk
2,2671214
2,2671214
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
|
show 1 more comment
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
1
1
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
Got it. Another way to do the argument is to first show that $H^2((-n, n))= 0$, which follows since $H^1((-n, n)) = 2n$, hence $H^alpha((-n, n)) = 0$ for all $alpha > 1$. Now $H^2(mathbf{R}) = lim_n H^2((-n,n)) = 0$, by continuity of measure and measurability of Borels. This is not as elementary though, of course.
$endgroup$
– Drew Brady
Jan 8 at 15:10
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
actually what goes wrong in your argument if you use this cover for $H^1$, though? I'm confused, don't you get $H^1_delta(mathbf{R}) leq delta$?
$endgroup$
– Drew Brady
Jan 8 at 15:19
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
@DrewBrady, thanks for pointing this out. Actually I noticed that also, that I'm not really using the fact that we deal with $2d$ measure. I've now revised the argument. Taking small rectangles, sum of whose lengths diverge, but converge with when in squared summation. This will ensure that we cover $mathbb{R}$ using arbitrarily small $2d$ area.
$endgroup$
– Hayk
Jan 8 at 15:30
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
I don't understand what was wrong previously, though. Was it not a cover of $mathbf{R}$? In particular, let $0 < eta < 1$, and let $B_n = B(r_n, eta delta 2^{-n})$. Then $mathrm{diam}(B_n) = eta delta 2^{1-n} < delta$. Hence $H^1_delta(mathbf{R}) leq sum_{n} eta delta 2^{-n} leq eta delta$. Taking $eta downarrow 0$, this gives $H^1_delta(mathbf{R}) = 0$, for all $delta$. But this clearly can't be correct.
$endgroup$
– Drew Brady
Jan 8 at 15:31
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
$begingroup$
@DrewBrady in the argument with balls of radius $delta 2^{-n}$ we are simple not covering $mathbb{R}$. Indeed, consider intersections of these balls with $mathbb{R}$. For the $n$-th ball, the intersection is an interval of length $delta 2^{-n}$. Now taking the sum over all these intervals (which do have overlaps) converges, and hence cannot cover $mathbb{R}$ whose length is infinite.
$endgroup$
– Hayk
Jan 8 at 15:34
|
show 1 more comment
